\newproblem{lay:1_9_23}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.9.23}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Nov. 11th, 2013} \\}{}

  % Problem statement
	For each of the following statements determine if they are True or False. Justify your answer.
	\begin{enumerate}
		\item A linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is completely determined by its effect on the columns of the 
		      matrix $n\times n$ identity matrix.
		\item If $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ rotates vectors about the origin through an angle $\phi$, then $T$ is a linear transformation.
		\item When two linear transformations are performed one after another, the combined effect may not always be a linear transformation.
		\item A mapping $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is onto $\mathbb{R}^m$ if every vector $\mathbf{x}$ in $\mathbb{R}^n$ maps onto some
		      vector in $\mathbb{R}^m$.
		\item If $A$ is a $3\times 2$ matrix, then the transformation $\mathbf{x}\rightarrow A\mathbf{x}$ cannot be one-to-one.
	\end{enumerate}
}{
  % Solution
	\begin{enumerate}
		\item True. Since the columns of $I_n$ are a basis of $\mathbb{R}^n$, then any vector in $\mathbb{R}^n$ can be expressed as:
		      \begin{center}
						$\mathbf{x}=x_1\mathbf{e}_1+...+x_n\mathbf{x}_n$
					\end{center}
					Since $T$ is a linear transformation, then
		      \begin{center}
						$T(\mathbf{x})=T(x_1\mathbf{e}_1+...+x_n\mathbf{e}_n)=x_1T(\mathbf{e}_1)+...+x_nT(\mathbf{e}_n)$
					\end{center}
					That is, to calculate $T(\mathbf{x})$ we only need to know how to transform the columns of the identity matrix of size $n\times n$.
		\item True. As shown in Example $1.9.3$, such a rotating transformation can be expressed as
		      \begin{center}
						$T(\mathbf{x})=\begin{pmatrix}\cos(\phi) & -\sin(\phi) \\ \sin(\phi) & \cos(\phi)\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}$
					\end{center}
					That is, it is of the form $T(\mathbf{x})=A\mathbf{x}$, that is a linear transformation.
		\item False. The composition of two linear transformations is also a linear transformation as shown below: \\
					Let $T_1:\mathbb{R}^n\rightarrow \mathbb{R}^m$ and $T_2:\mathbb{R}^m\rightarrow \mathbb{R}^p$ be two linear transformations, and
					$T_{12}:\mathbb{R}^n\rightarrow\mathbb{R}^p$ be defined as $(T_1\circ T_2)(\mathbf{x})=T_2(T_1(\mathbf{x}))$
		      \begin{itemize}
						\item We need to show that for all $\mathbf{u},\mathbf{v}\in\mathbb{R}^n$, it is verified that $T_{12}(\mathbf{u}+\mathbf{v})=T_{12}(\mathbf{u})+T_{12}(\mathbf{v})$\\
						     \begin{center}
										$\begin{array}{rcll}
											T_{12}(\mathbf{u}+\mathbf{v})&=&T_2(T_1(\mathbf{u}+\mathbf{v}))&\text{By definition} \\
											  &=&T_2(T_1(\mathbf{u})+T_1(\mathbf{v})) & T_1 \text{ is a linear transformation} \\
											  &=&T_2(T_1(\mathbf{u}))+T_2(T_1(\mathbf{v})) & T_2 \text{ is a linear transformation} \\
											  &=&T_{12}(\mathbf{u})+T_{12}(\mathbf{v}) & \text{By definition} \\
										\end{array}$
								 \end{center}
					\end{itemize}
		\item False. For instance the mapping $T:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ given by $T(x_1,x_2)=(x_1,x_2,0)$ produces a vector in $\mathbb{R}^3$ for every vector
		      in $\mathbb{R}^2$. However, the transformation is not onto $\mathbb{R}^3$ because there are vectors in this space that are not coming from any input vector
					(for instance, the vector $(0,0,1)$ is not the image of any of the vectors in $\mathbb{R}^2$).
					
		\item False. Consider the matrix $A=\begin{pmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{pmatrix}$. The transformation $T$ is, then, defined as $T(x_1,x_2)=(x_1,x_2,0)$.
		      Let us analyze how many input vectors map onto each vector $\mathbf{y}$ in $\mathbb{R}^3$. For doing so, let us analyze the equation system $A\mathbf{x}=\mathbf{y}$
					\begin{center}
					   $\left(\begin{array}{rr|r} 1 & 0 & y_1 \\ 0 & 1 & y_2 \\ 0 & 0 & y_3 \end{array}\right)$
					\end{center}
					This equation system has no solution if $y_3\neq 0$, and a unique solution if $y_3=0$ (the unique solution is $x_1=y_1$ and $x_2=y_2$). Therefore, the transformation
					is one-to-one.
	\end{enumerate}
}
\useproblem{lay:1_9_23}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
